The question is the following: Can one create two nonidentical loaded 6-sided dice such that when one throws with both dice and sums their values the probability of any sum (from 2 to 12) is the same. I said nonidentical because its easy to verify that with identical loaded dice its not possible.

Just a side note: A famous example of the generating function techniques described in the answers is the derivation of the "Sicherman dice", two unequal dice with the same distribution for the sum as a pair of ordinary six-sided dice. See en.wikipedia.org/wiki/Sicherman_dice.
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Hans LundmarkOct 6 '10 at 20:31

4

I just want to note that I am one of the people voting to close. My concern is that this is a very standard exercise on how to use generating functions to work with probability; I was assigned it as an undergrad and I'm sure I will assign it in turn.
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David SpeyerOct 6 '10 at 21:25

I hadn't seen it, which is why I answered it. (I almost voted to close, actually, until I realized this was not the Sicherman dice problem.)
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Michael LugoOct 6 '10 at 21:44

Fair enough, and your answer is well written. This sort of thing is always going to grey areas, but I didn't like that there were three votes to close with no explanation.
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David SpeyerOct 6 '10 at 21:59

5 Answers
5

You can write this as a single polynomial equation
$$p(x)q(x)=\frac1{11}(x^2+x^3+\cdots+x^{12})$$
where $p(x)=p_1x+p_2x^2+\cdots+p_6x^6$ and similarly for $q(x)$.
So this reduces to the question of factorizing $(x^2+\cdots+x^{12})/11$
where the factors satisfy some extra conditions (coefficients positive,
$p(1)=1$ etc.).

Then the coefficient of $x^n$ in $P(x) Q(x)$ is exactly the probability that the sum of your two dice is $n$. As Robin Chapman points out, you want to know if it's possible to have

$$ P(x) Q(x) = (x^2 + \cdots + x^{12})/11 $$

where $P$ and $Q$ are both sixth-degree polynomials with positive coefficients and zero constant term.

For simplicity, I'll let $p(x) = P(x)/x, q(x) = Q(x)/x$. Then we want

$$ p(x) q(x) = (1 + \cdots + x^{10})/11 $$

where $p$ and $q$ are now fifth-degree polynomials. We can rewrite the right-hand side to get

$$ p(x) q(x) = {(x^{11}-1) \over 11(x-1)} $$

or

$$ 11 (x-1) p(x) q(x) = x^{11} - 1. $$

The roots of the right-hand side are the eleventh roots of unity. Therefore the roots of $p$ must be five of the eleventh roots of unity which aren't equal to one, and the roots of $q$ must be the other five.

But the coefficients of $p$ and $q$ are real, which means that their roots must occur in complex conjugate pairs. So $p$ and $q$ must have even degree! Since five is not even, this is impossible.

(This proof would work if you replace six-sided dice with any even-sided dice. I suspect that what you want is impossible for odd-sided dice, as well, but this particular proof doesn't work.)

I've heard this brainteaser before, and usually its phrased that 2-12 must come up equally likely (no comment about other sums). With this formulation (or interpretation) it becomes possible. Namely, {0,0,0,6,6,6} and {1,2,3,4,5,6}. In this case, you can also generate the sum 1, but 2-12 are equally likely (1-12 are equally likely). Without allowing for other sums I do suspect its impossible (and looks like proofs have been given). I arrived at this answer by noting we are asking for equal probability for 11 events that come from 36 (6*6) possible outcomes, which immediately seems unlikely. However equal probability for 12 events from 36 outcomes is far more manageable :)

You can't even solve this with two-sided dice. Consider two dice with probabilities p and q of rolling 1, and probabilities (1-p) and (1-q) of rolling 2. The probability of rolling a sum of 2 is pq, and the probability of rolling a sum of 4 is (1-p)(1-q). These are equal only if p=(1-q). Hence they are equal to one third only if p satisfies the quadratic equation p(1-p) = 1/3. Since this has no real roots, it cannot be done. This logic extends to multisided dice.